dc123 wrote:

|x| < 1?

1) |x+1| = 2|x-1|

2) |x-3| does not equal 0

Therefore we are looking to see if -1<x<1????

Can someone plz help. I dont know why its C

You can do these questions in less than a minute if you understand the approach I discuss here:

http://www.veritasprep.com/blog/2011/01 ... edore-did/On to your question.

Is |x| < 1?

Yes, essentially this just asks you whether x lies between -1 and 1. |x| represents the distance from 0 which will be less than 1 if x lies within -1 < x < 1.

Statement 1: |x+1| = 2|x-1|

Draw the number line. This equation says that distance from -1 is twice the distance from 1. At which point is the distance from -1, twice the distance from 1? If you split the distance (2 units) between -1 and 1 into 3 parts, 1 part away from 1 and 2 parts away from -1 will be the point 1/3. The diagram below will show you this situation. Similarly the distance between -1 and 1 is 2 so if you go 2 units further to the right of 1, you get the point 3 which will be 4 units away from -1 i.e. twice the distance from 1.

Attachment:

Ques6.jpg [ 4.74 KiB | Viewed 1055 times ]
x could be 1/3 or 3. Hence x may or may not lie between -1 and 1. Not sufficient.

Statement 2: |x-3| does not equal 0

This statement just says that x is not 3. It is not sufficient alone.

Both together, we get that x must be either 3 or 1/3 and it is not 3. Then x must be 1/3 and must lie between -1 and 1. Sufficient.

Answer (C)

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